我应该继续缩短我的equals和hashcode函数吗？

当我有这样的课程

public class MyClass {
    private ClassA classA;
    private ClassB classB;
}

我的equals和hascode函数经常会像这样混乱

@Override
public int hashCode() {
    final int prime = 31;
    int result = 1;
    result = prime * result + ((classB == null) ? 0 : classB.hashCode());
    result = prime * result + ((classA == null) ? 0 : classA.hashCode());
    return result;
}

@Override
public boolean equals(Object obj) {
    if (this == obj)
        return true;
    if (obj == null)
        return false;
    if (getClass() != obj.getClass())
        return false;
    MyClass other = (MyClass) obj;
    if (classB == null) {
        if (other.classB != null)
            return false;
    } else if (!classB.equals(other.classB))
        return false;
    if (classA == null) {
        if (other.classA != null)
            return false;
    } else if (!classA.equals(other.classA))
        return false;
    return true;
}

所以我开始忽略有人会将苹果与香蕉进行比较.我的应用程序中的HashSets或HashMaps总是包含相同的对象类型...我从未创建过List并开始将Integer与String混合.

@Override
public int hashCode() {
    return classB.hashCode();
}

@Override
public boolean equals(Object obj) {
    MyClass other = (MyClass) obj;
    return other.classB.equals(this.classB) && other.classA.equals(this.classA);
}

忽视不常见的情况并抛出异常是不好或好的做法？我认为当比较完整的不同类时,这通常是一个错误.

允许equals()或hashCode()抛出异常是非常糟糕的做法(我建议您阅读Effective Java的详细信息).

此外,您的方法不必要地复杂.从Java 7开始,这几乎就是编写这些方法的规范方法:

@Override
public boolean equals(Object o) {
    if (this == o) {
        return true;
    }else if (o instanceof MyClass) {
        // some will argue that the line above should have a
        // .getClass() check instead of an instanceof check.
        // those people also need to read Effective Java :-)
        MyClass that = (MyClass) o;
        return Objects.equals(this.classA, that.classA)
            && Objects.equals(this.classB, that.classB)
    } else {
        return false;
    } 
}

@Override
public int hashCode() {
    return Objects.hash(classA, classB);
}